Answer the following

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example is given below please follow the method in the solution

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Solutions: a. v after Is d. t from origin to maximum height use: vS = vi - gt Note: An object thrown upward will reach a certain point (maximum vy=20 m/s-(9.8 m/s2) (1.0s) height) where it will momentarily vs = 20 m/s - 9.8 m/s stop and then start to move vy = 10.2 m/s downward. At this point, vs= 0. b. v after 2s Use: v = v1 - gt use: v/ = v1 - gt Manipulating the equation, vy= 20 m/s - (9.8 m/s2) (2.0s) vy = 20 m/s - 19.6 m/s t = V1 - Up vs = 0.4 m/s t = 20 m/s - 0 c. d after Is 9.8 m/s2 use: d = vit - 9 t = 2.04 s d = 20m/s(1.0s) - (9.8 m/s2)(1.0s)2 e. d from origin to maximum height 2 use: d = 20m - 4.9 m d = vit - 9tz d = 15.1 m d = 20m/s (2.04 s) - (9.8 m/s2)(2.04s)2 (positive sign denotes that the object is 2 moving upward) d = 40.8 m - 20.40 m d = 20.4 mSample Problem 1. A stone is dropped from the top of a building and falls freely from rest. What is its position (height) and its velocity after 1.0s, 2.0s, and 3.0s? To find for the position(height), we will c. Find h after 3.0s use: h = 0(3.0s) - (9.8 m/s2)(3.0s)2 h = vit- g 2 2 h = 0-88.2 m Know that initial velocity (vi) = 0 2 a. Find h after 1.0s h = - 44.1 m or 44.1 m downward h = 0(1.0s) - (9.8 m/s2)(1.0s)2 To find for velocity (vr), we will use: 2 v/ = v1 - gt h =0 -9.8 m 2 a. Find vs after 1.0s h = - 4.9 m vs = 0- (9.8 m/s2) (1.0s) (negative sign denotes downward direction) vy = - 9.8 m/s b. Find h after 2.0s b. Find vy after 2.0s h = 0(2.0s) - (9.8 m/s?)(2.0s)2 v= 0 - (9.8 m/s2) (2.0s) vy = - 19.6 m/s h =0-39.2m c. Find vy after 3.0s 2 vs=0 - (9.8 m/s2) (.0s) h = - 19.6 m or 19.6 m downward vf= - 29.4 m/s Sample Problem 2. A boy threw a ball upward with an initial velocity of 20 m/s and was able to catch it before it reached the ground on its return. a. What was the ball's velocity after Is? b. What was the ball's velocity after 2s? c. What was its displacement after Is? d. How long did it take the ball to reach its maximum height? e. How far was the maximum height from the starting point? Given: vi = 20 m/s Find: a. v after Is b. v after 5s c. d after Is d. t from origin to maximum height e. d from origin to maximum heightActivity 2: Tell Me! Directions: Give what is asked in each item based on the situation. Write your answer on your answer sheet. A ball is thrown vertically upward. a. What is the acceleration of the ball after I s? b. What is the acceleration of the ball at its maximum height? c. What is the velocity of the ball at the top of its path? d. If the ball is thrown with an initial velocity of 50 m/s, what is the final velocity of the ball at a point at the same level as when it was thrown? e. What force was exerted on the ball 1 second after it was thrown? f. What exerts this force? g. What is the acceleration of the ball 10 seconds after being thrown up? h. If the ball took 5 s to reach its maximum height, how long will the ball go back to where it was initially launched? i. What is the acceleration of the ball at the top of its path? j. What is the speed of the ball at the top of its path? It's now time to check your learning from this module. Complete the statement